PSEB 12th Class Maths Solutions Chapter 7 Integrals Ex 7.2

Punjab State Board PSEB 12th Class Maths Book Solutions Chapter 7 Integrals Ex 7.2 Textbook Exercise Questions and Answers.

PSEB Solutions for Class 12 Maths Chapter 7 Integrals Ex 7.2

Direction (1 – 37): Integrate the following functions.

Question 1.
\(\frac{2 x}{1+x^{2}}\)
Solution.
∫ \(\frac{2 x}{1+x^{2}}\) dx
Let 1 + x² = t
On differentiating w.rt. x, we get
2x dx = dt
⇒ x dx = \(\frac{d t}{2}\)
⇒ ∫ \(\frac{2 x}{1+x^{2}}\) dx = ∫ \(\frac{1}{t}\) dt
= log |t| + C
= log |1 + x²| + C

Question 2.
\(\frac{(\log x)^{2}}{x}\)
Sol.
∫ \(\frac{(\log x)^{2}}{x}\) dx
Let log x = t
On differentiating w.r.t. x, we get
\(\frac{1}{x}\) dx = dt
⇒ dx = x dt
⇒ ∫ \(\frac{(\log x)^{2}}{x}\) dx = ∫ t² dt
= \(\frac{t^{3}}{3}\) + C
= \(\frac{(\log x)^{3}}{3}\) + C

Question 3.
\(\frac{1}{x+x \log x}\)
Solution.
∫ \(\frac{1}{x+x \log x}\) dx = ∫ \(\frac{1}{x(1+\log x)}\) dx
Let 1 + log x = t
On differentiating w.r.t. x, we get
\(\frac{1}{x}\) dx = dt
⇒ dx = x dt
⇒ ∫ \(\frac{1}{x+x \log x}\) dx = ∫ \(\frac{1}{t}\) dt
= log |t| + C
= log |1 + log x| + C

Question 4.
sin x . sin (cos x)
Solution.
∫ sin x . sin(cos x) dx
Let cos x = t
On differentiating w.r.t. x, we get
– sin x dx = dt
⇒ ∫ sin x . sin (cos x) dx = – ∫ sin t dt
= – [- cos t] + C
= cos t + C = cos (cos x) + C

Question 5.
sin(ax + b) cos(ax + b)
Solution.
∫ sin(ax + b) cos(ax + b) dx
= – \(\frac{1}{2}\) ∫ 2 sin (ax + b) cos(ax + b) dx
= – \(\frac{1}{2}\) ∫ sin (2ax + 2b) dx
Let 2ax + 2b = t
On differentiating w.r.t. x, we get
2a dx = dt
∴ I = \(\frac{1}{2}\) ∫ sin t . \(\frac{d t}{2 a}\)

= \(\frac{1}{4 a}\) ∫ sin t dt

= – \(\frac{1}{4 a}\) cos t + C

= – \(\frac{1}{4 a}\) cos (2ax + 2) + C

Question 6.
\(\sqrt{a x+b}\)
Solution.
∫ \(\sqrt{a x+b}\) dx
Let ax + b = t
On differentiating w.r.t. x, we get
a dx = dt
∴ dx = \(\frac{1}{a}\) dt
⇒ ∫ \((a x+b)^{\frac{1}{2}} d x=\frac{1}{a} \int t^{\frac{1}{2}} d t\)

= \(\frac{1}{a}\left(\frac{t^{\frac{3}{2}}}{\frac{3}{2}}\right)\)

= \(\frac{2}{3 a}(a x+b)^{\frac{3}{2}}\) + C

Question 7.
x \(\sqrt{x+2}\)
Solution.
∫ x \(\sqrt{x+2}\) dx
Let (x + 2) = t
On differentiating w.r.t. x, we get
dx = dt

Question 8.
x \(\sqrt{1+2 x^{2}}\)
Solution.
∫ x \(\sqrt{1+2 x^{2}}\) dx
Let 1 + 2x² = t
On differentiating w.r.t. x, we get
4x dx = dt
∫ x \(\sqrt{1+2 x^{2}}\) dx = \(\int \frac{\sqrt{t} d t}{4}=\frac{1}{4} \int t^{\frac{1}{2}} d t\)

= \(\frac{1}{4}\left(\frac{t^{\frac{3}{2}}}{\frac{3}{2}}\right)+C=\frac{1}{6}\left(1+2 x^{2}\right)^{\frac{3}{2}}+C\)

Question 9.
(4x + 2) \(\sqrt{x^{2}+x+1}\)
Solution.
∫ (4x + 2) \(\sqrt{x^{2}+x+1}\) dx
Let x² + 2x + 1 = t
On differentiating w.r.t. x, we get
(2x + 1) dx = dt
∫ (4x + 2) \(\sqrt{x^{2}+x+1}\) dx = ∫ 2√t dt = 2 ∫ √t dt
= 2 \(\left(\frac{t^{\frac{3}{2}}}{\frac{3}{2}}\right)\) + C

= \(\frac{4}{3}\left(x^{2}+x+1\right)^{\frac{3}{2}}\) + C

Question 10.
\(\frac{1}{x-\sqrt{x}}\)
Solution.
∫ \(\frac{1}{x-\sqrt{x}}\) dx = ∫ \(\frac{1}{\sqrt{x}(\sqrt{x}-1)}\) dx
Let (√x – 1) = t
On differentiating w.r.t. x, we get
∴ \(\frac{1}{2 \sqrt{x}}\) dx = dt
⇒ dx = 2√x dt
⇒ ∫ \(\frac{1}{2 \sqrt{x}}\) dx = ∫ \(\frac{1}{\sqrt{x} t} 2 \sqrt{x} d t=\int \frac{2}{t} d t\)
= 2 log |t| + C
= 2 log |\(\sqrt{x}-1\)| + C

Question 11.
\(\frac{x}{\sqrt{x+4}}\), x > 0
Solution.
∫ \(\frac{x}{\sqrt{x+4}}\) dx
Let x + 4 = t
On differentiating w.r.t. x, we get
dx = dt

Question 11.
\(\frac{x}{\sqrt{x+4}}\), x > 0
Solution.
∫ \(\frac{x}{\sqrt{x+4}}[latex] dx
Let x + 4 = t
On differentiating w.r.t. x, we get
dx = dt

= [latex]\frac{2}{3}(x+4)^{\frac{1}{2}}\) (x + 4 – 12) + C

= \(\frac{2}{3} \sqrt{x+4}\) (x – 8) + C

Question 12.
\(\left(x^{3}-1\right)^{\frac{1}{3}} x^{5}\)
Solution.
∫ \(\left(x^{3}-1\right)^{\frac{1}{3}} x^{5}\) dx
Let x³ – 1 = t
On differentiating w.r.t. x, we get
3x² dx = dt
⇒ dx = \(\frac{d t}{3 x^{2}}\)
⇒ ∫ \(\left(x^{3}-1\right)^{\frac{1}{3}} x^{5}\) dx = ∫ (x³ – 1)^{\(\frac{4}{4}\)} x³ x² dx

Question 13.
\(\frac{x^{2}}{\left(2+3 x^{3}\right)^{3}}\)
Solution.
∫ \(\frac{x^{2}}{\left(2+3 x^{3}\right)^{3}}\) dx
On differentiating w.r.t. x, we get
9x² dx = dt
⇒ dx = \(\frac{d t}{9 x^{2}}\)
∫ \(\frac{x^{2}}{\left(2+3 x^{3}\right)^{3}}\) dx = \(\frac{1}{9} \int \frac{d t}{(t)^{3}}=\frac{1}{9}\left[\frac{t^{-3+1}}{-3+1}\right]+C\)

= – \(\frac{1}{18}\left(\frac{1}{t^{2}}\right)\) + C

= – \(\frac{1}{18\left(2+3 x^{3}\right)^{2}}\) + C

Question 14.
\(\frac{1}{x(\log x)^{m}}\), x > 0, m ≠ 1
Solution.
∫ \(\frac{1}{x(\log x)^{m}}\) dx
Let log x = t
On differentiating w.r.t. x, we get
\(\frac{1}{x}\) dx = dt
⇒ dx = x dt
∴ ∫ \(\frac{1}{x(\log x)^{m}}\) dx = ∫ \(\frac{d t}{(t)^{m}}\)

= \(\left(\frac{t^{-m+1}}{-m+1}\right)\) + C

= \(\frac{(\log x)^{1-m}}{(1-m)}\) + C

Question 15.
\(\frac{x}{9-4 x^{2}}\)
Solution.
∫ \(\frac{x}{9-4 x^{2}}\) dx
Let 9 – 4x² = t
On differentiating w.r.t. x, we get
– 8x dx = dt
⇒ dx = \(\frac{d t}{-8 x}\)
∫ \(\frac{x}{9-4 x^{2}}\) dx = \(\frac{-1}{8}\) ∫ \(\frac{1}{t}\) dt
= \(\frac{-1}{8}\) log |t| + C
= \(\frac{-1}{8}\) log |9 – 4x²| + C

Question 16.
e^{2x +3}
Solution.
∫ e^{2x + 3} dx
Let 2x + 3 = t
On differentiating w.r.t. x, we get
2 dx = dt
⇒ dx = \(\frac{1}{2}\) dt
∴ ∫ e^{2x + 3} dx = \(\frac{1}{2}\) ∫ e^t dt
= \(\frac{1}{2}\) (e^t) + C
= \(\frac{1}{2}\) e^{(2x + 3)} + C

Question 18.
\(\frac{e^{\tan ^{-1} x}}{1+x^{2}}\)
Solution.
∫ \(\frac{e^{\tan ^{-1} x}}{1+x^{2}}\) dx
Let tan^-1 x = t
⇒ \(\frac{1}{1+x^{2}}\) dx = dt
⇒ dx = (1 + x²) dt
∴ ∫ \(\frac{e^{\tan ^{-1} x}}{1+x^{2}}\) dx = ∫ e^t dt
= e^t + C
= e^tan-1 x + C

Question 19.
\(\frac{e^{2 x}-1}{e^{2 x}+1}\)
Solution.

= log |t| + C
= log |e^x + e^-x| + C.

Question 20.
\(\frac{e^{2 x}-e^{-2 x}}{e^{2 x}+e^{-2 x}}\)
Solution.
∫ \(\frac{e^{2 x}-e^{-2 x}}{e^{2 x}+e^{-2 x}}\) dx
Let e^2x + e^-2x = t
⇒ (2 e^2x – 2 e^-2x) dx = dt
⇒ dx = \(\frac{d t}{2\left(e^{2 x}-e^{-2 x}\right)}\)

∫ \(\frac{e^{2 x}-e^{-2 x}}{e^{2 x}+e^{-2 x}}\) dx = ∫ \(\frac{d t}{2 t}\)

= \(\frac{1}{2}\) ∫ \(\frac{1}{t}\) dt

= \(\frac{1}{2}\) log |t| + C

= \(\frac{1}{2}\) log |e^2x + e^-2x| + C.

Question 21.
tan² (2x – 3)
Solution.
∫ tan² (2x – 3) dx = ∫ sec² (2x – 3) dx – ∫ 1 dx
Let 2x – 3 = t
⇒ 2 dx = dt
⇒ dx = \(\frac{1}{2}\) dt
⇒ ∫ sec² (2x – 3) dx – ∫ 1 dx = \(\frac{1}{2}\) ∫ (sec² t dt – ∫ 1 dx
= \(\frac{1}{2}\) tan t – x + C
= \(\frac{1}{2}\) tan (2x – 3) – x + C

Question 22.
sec² (7 – 4x)
Solution.
∫ sec² (7 – 4x) dx
Let 7 – 4x = t
⇒ – 4 dx = dt
⇒ dx = \(\frac{d t}{-4}\)

∴ ∫ sec² (7 – 4x) dx = \(\frac{-1}{4}\) ∫ sec² t dt
= \(\frac{-1}{4}\) (tan t) + C
= \(\frac{-1}{4}\) tan (7 – 4x) + C

Question 23.
\(\frac{\sin ^{-1} x}{\sqrt{1-x^{2}}}\)
Solution.
∫ \(\frac{\sin ^{-1} x}{\sqrt{1-x^{2}}}\) dx
Let sin^-1 x = t

Question 24.
\(\frac{2 \cos x-3 \sin x}{6 \cos x+4 \sin x}\)
Solution.
∫ \(\frac{2 \cos x-3 \sin x}{6 \cos x+4 \sin x}\) dx = ∫ \(\frac{2 \cos x-3 \sin x}{2(3 \cos x+2 \sin x)}\) dx

Let 3 cos x + 2 sin x = t

⇒ (- 3 sin x + 2 cos x) dx = dt

⇒ dx = \(\frac{d t}{2 \cos x-3 \sin x}\)

∴ ∫ \(\frac{2 \cos x-3 \sin x}{6 \cos x+4 \sin x}\) dx = ∫ \(\frac{d t}{2 t}\) + \(\frac{1}{2}\) ∫ \(\frac{1}{t}\) dt

= \(\frac{1}{2}\) log |t| + C

= \(\frac{1}{2}\) log |2 sin x + 3 cos x| + C

Question 25.
\(\frac{1}{\cos ^{2} x(1-\tan x)^{2}}\)
Solution.
∫ \(\frac{1}{\cos ^{2} x(1-\tan x)^{2}}\) dx = ∫ \(\frac{\sec ^{2} x}{(1-\tan x)^{2}}\) dx
Let (1 – tan x) = t
⇒ – sec² x dx = dt
⇒ dx = \(\frac{d t}{-\sec ^{2} x}\)

∫ \(\frac{\sec ^{2} x}{(1-\tan x)^{2}}\) dx = ∫ \(\frac{-d t}{t^{2}}\)
= – ∫ t² dt

= – \(\left(\frac{-t^{-2+1}}{-2+1}\right)\) + C

= \(\frac{1}{t}+C=\frac{1}{(1-\tan x)}\) + C

Question 26.
\(\frac{\cos \sqrt{x}}{\sqrt{x}}\)
Solution.
∫ \(\frac{\cos \sqrt{x}}{\sqrt{x}}\) dx
Let √x = t
⇒ \(\frac{1}{2 \sqrt{x}}\) dx = dt
⇒ dx = 2 √x dt
∫ \(\frac{\cos \sqrt{x}}{\sqrt{x}}\) dx = 2 ∫ cos t dt
= 2 sin t + C
= 2 sin √x + C

Question 27.
\(\sqrt{\sin 2 x}\) cos 2x
Solution.
∫ \(\sqrt{\sin 2 x}\) cos 2x dx
Let sin 2x = t
⇒ 2 cos 2x dx = dt
⇒ dx = \(\frac{d t}{2 \cos 2 x}\)
∴ ∫ \(\sqrt{\sin 2 x}\) cos 2x dx = \(\frac{1}{2}\) ∫ (t)^{\(\frac{1}{2}\)} dt

= \(\frac{1}{2}\left(\frac{t^{\frac{3}{2}}}{\frac{3}{2}}\right)\)

= \(\frac{1}{3} t^{\frac{3}{2}}\) + C

= \(\frac{1}{3}(\sin 2 x)^{\frac{3}{2}}\) + C

Question 28.
\(\frac{\cos x}{\sqrt{1+\sin x}}\)
Solution.
∫ \(\frac{\cos x}{\sqrt{1+\sin x}}\) dx
Let 1 + sin x = t
⇒ cos x dx = dt
⇒ dx = \(\frac{d t}{\cos x}\)

∫ \(\frac{\cos x}{\sqrt{1+\sin x}}\) dx = \(\frac{d t}{\sqrt{t}}\)

= \(\frac{t^{\frac{1}{2}}}{\frac{1}{2}}\) + C

= 2√t + C

= 2 \({\sqrt{1+\sin x}}\) + C

Question 29.
cot x log sin x
Solution.
∫ cot x log sin x dx
Let log sin x = t .
⇒ \(\frac{1}{\sin x}\) . cos x dx = dt
⇒ cot x dx = dt
⇒ dx = \(\frac{d t}{\cot x}\)
∴ ∫ cot x log sin x dx = ∫ t dt
= \(\frac{t^{2}}{2}\) + C
= \(\frac{1}{2}\) (log sin x)² + C

Question 30.
\(\frac{\sin x}{1+\cos x}\)
Solution.
∫ \(\frac{\sin x}{1+\cos x}\) dx
Let 1 + cos x = t
⇒ – sin x dx = dt
⇒ dx = \(\frac{d t}{-\sin x}\)
∴ ∫ \(\frac{\sin x}{1+\cos x}\) dx = ∫ – \(\frac{d t}{t}\)
= – log |t| + C
= – log |1 + cos x| + C
= log \(\frac{1}{|1+\cos x|}\) + C

Question 31.
\(\frac{\sin x}{(1+\cos x)^{2}}\)
Solution.
∫ \(\frac{\sin x}{(1+\cos x)^{2}}\) dx
Let 1 + cos x = t
⇒ – sin x dx = dt
⇒ dx = \(=\frac{d t}{-\sin x}\)
∴ ∫ \(\frac{\sin x}{(1+\cos x)^{2}}\) dx = ∫ – \(\frac{d t}{t^{2}}\)

= ∫ t^-2 dt

= \(\frac{-t^{-2+1}}{-2+1}\) + C

= \(\frac{1}{t}\) + C

= \(\frac{1}{1+\cos x}\) + C

Question 32.
\(\frac{1}{1+\cot x}\)
Solution.

Question 33.
\(\frac{1}{1-\tan x}\)
Solution.

Question 34.
\(\frac{\sqrt{\tan x}}{\sin x \cos x}\)
Solution.
Let I = \(\frac{\sqrt{\tan x}}{\sin x \cos x}\) dx

= ∫ \(\frac{\sqrt{\tan x} \times \cos x}{\sin x \cos x \times \cos x}\) dx

= ∫ \(\frac{\sqrt{\tan x}}{\tan x \cos ^{2} x}\) dx

= ∫ \(\frac{\sec ^{2} x}{\sqrt{\tan x}}\) dx

Let x = tan t
⇒ sec² x dx = dt
⇒ dx = \(\frac{d t}{\sec ^{2} x}\)

∴ I = ∫ \(\frac{d t}{\sqrt{t}}\)

= ∫ t^{\(-\frac{1}{2}\)}

= 2√t + C

= 2√tan x + C

Question 35.
\(\frac{(1+\log x)^{2}}{x}\)
Solution.
∫ \(\frac{(1+\log x)^{2}}{x}\) dx
Let 1 + log x = t
⇒ \(\frac{1}{x}\)dx = dt
⇒ dx = x dt
∴ ∫ \(\frac{(1+\log x)^{2}}{x}\) dx = ∫ t² dt
= \(\frac{t^{3}}{3}\) + C
= \(\frac{(1+\log x)^{3}}{3}\) + C

Question 36.
\(\frac{(x+1)(x+\log x)^{2}}{x}\)
Solution.
∫ \(\frac{(x+1)(x+\log x)^{2}}{x}\) dx = ∫ \(\frac{x+1}{x}\) (x + log x)² dx

= ∫ (1 + \(\frac{1}{x}\)) (x + log x)² dx

Let x + log x = t
⇒ (1 + \(\frac{1}{x}\)) dx = dt

∴ ∫ (1 + \(\frac{1}{x}\)) (x + log x)² dx = ∫ t² dt

= \(\frac{t^{3}}{3}\) + C

= \(\frac{1}{3}\) (x + log x)³ + C

Question 37.
\(\frac{x^{3} \sin \left(\tan ^{-1} x^{4}\right)}{1+x^{8}}\)
Solution.
Let I = ∫ \(\frac{x^{3} \sin \left(\tan ^{-1} x^{4}\right)}{1+x^{8}}\) dx

Let tan^-1 x⁴ = t
⇒ \(\frac{1}{1+x^{8}}\) . 4x³ = \(\frac{d t}{d x}=\)

⇒ dx = \(\frac{\left(1+x^{8}\right)}{4 x^{3}}\) dt

∴ I = ∫ \(\frac{x^{3} \sin t}{\left(1+x^{8}\right)} \cdot \frac{1+x^{8} d t}{4 x^{3}}\)

= \(\frac{1}{4}\) ∫ sin t dt

= – \(\frac{1}{4}\) cos t + C

Direction (38 – 39) : Choose the correct answer.

Question 38.
∫ \(\frac{10 x^{9}+10^{x} \log _{e} 10}{x^{10}+10^{x}}\) dx equals
(A) 10^x – x¹⁰ + C

(B) 10^x + x¹⁰ + C

(C) (10^x – x¹⁰)^-1 + C

(D) log (10^x + x¹⁰) + C
Solution.
∫ \(\frac{10 x^{9}+10^{x} \log _{e} 10}{x^{10}+10^{x}}\) dx
Let x¹⁰ + 10^x = t
⇒ dx = \(\frac{d t}{10 x^{9}+10^{x} \log _{e} 10}\)
∴ ∫ \(\frac{10 x^{9}+10^{x} \log _{e} 10}{x^{10}+10^{x}}\) dx = ∫ \(\frac{dt}{t}\)
= log t + C
= log (10^x + x¹⁰) + C
Hence, the correct answer is (D).

Question 39.
∫ \(\frac{d x}{\sin ^{2} x \cos ^{2} x}\) ,
(A) tan x + cot x + C
(B) tan x – cot x + C
(C) tan x cot x + C
(D) tan x – cot 2x + C
Solution.

Hence, the correct answer is (B).